Establish convergence of the series: $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-...$

Question

Establish convergence of the series: $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-...$ The number of signs increases by one in each "block".

I have an idea. Group the series like this: $1-(\frac{1}{2}+\frac{1}{3})+(\frac{1}{4}+\frac{1}{5}+\frac{1}{6})-...$

We can show that $1, \frac{1}{2}+\frac{1}{3},\frac{1}{4}+\frac{1}{5}+\frac{1}{6},...$ converges to 0. I'm trying to use Dirichlet's Test. However, I'm not sure wether this sequence is decreasing.

Any idea? Or any other method to establish the convergence?

Answer 1

We can see that the general term of this series is $$a_n=(-1)^n\sum_{k=\frac{n(n+1)}{2}+1}^{\frac{n(n+1)}{2}+n+1}\frac{1}{k}$$ and we have $$H_n=\sum_{k=1}^n\frac{1}{k}=\log n+\gamma+\frac{1}{2n}+O(\frac{1}{n^2})$$ so \begin{align}|a_n|=H_{\frac{n(n+1)}{2}+n+1}-H_{\frac{n(n+1)}{2}+1}&=\log(1+\frac{2}{n})+\frac{1}{(n+1)(n+2)}-\frac{1}{n(n+1)}+O(\frac{1}{n^2})\\ &=\frac{2}{n}+O(\frac{1}{n^2})\end{align} hence we have $$a_n=\frac{2(-1)^n}{n}+O(\frac{1}{n^2})$$ which allows us to conclude the convergence of the series since it's sum of two convergent series, one by alternating series test and the other by comparaison with Riemann series.

Answer 2

If I'm not getting it wrong, your series is

$$1-\frac 1 2 -\frac 13+\frac 14+\frac 15+\frac 16-\frac 17-\frac 18-\frac 19-\frac 1{10}+++++------\dots$$

So you have $1$ plus, $2$ minuses, $3$ pluses, $4$ minuses, and so on.

We can write your series as $a_0+a_1+a_2+a_3+\dots$ where $$\begin{align} a_0&=(-1)^0 1 \\ a_1&=(-1)^1\sum_{k=2}^{3}\frac 1 k\\ a_2&=(-1)^2\sum_{k=4}^{6}\frac 1 k\\ a_3&=(-1)^3\sum_{k=7}^{10}\frac 1 k\\ \cdots &=\cdots \\ a_n&=(-1)^n\sum_{k=T_n+1}^{T_{n+1}}\frac 1 k\end{align}$$

Where $T_n=\frac{n(n+1)}{2}$ so it goes $1,3,6,10,\dots$

Now, we know that $$\sum_{k=1}^n \frac 1 k =\log n+\gamma+\frac 1 {2n}+O(n^{-2})$$

Thus

$$\eqalign{ & \sum\limits_{k = 1}^{{T_n}} {{1 \over k}} = \log n + \log \left( {n + 1} \right) - \log 2 + \gamma + O({n^{ - 2}}) \cr & \sum\limits_{k = 1}^{{T_{n + 1}}} {{1 \over k}} = \log \left( {n + 2} \right) + \log \left( {n + 1} \right) - \log 2 + \gamma + O({n^{ - 2}}) \cr} $$

Whence, after simplification

$$\sum\limits_{k = {T_n} + 1}^{{T_{n + 1}}} {{1 \over k}} = \log \left( {1 + {2 \over n}} \right) + O(n^{-2})$$

Recall that $$\log(1+x)=x+O(x^2)$$ so

$$\sum\limits_{k = {T_n} + 1}^{{T_{n + 1}}} {{1 \over k}} = \frac{2}{n} +O\left(\frac 1 {n^2}\right)$$

Since $$\sum (-1)^n \frac 1 n $$ and $$\sum n^{-2}$$ converge, so does your series.

Answer 3

You can use the alternating series test if you can prove that the sum of a block goes to zero. Block $n$ starts at $\dfrac 1{\dfrac {n(n-1)}2+1}$ and ends at $\dfrac 1{\frac {n(n+1)}2}$ and has $n$ terms. The sum is then less than $\dfrac n{\dfrac {n(n-1)}2}=\dfrac 2{n-1}$ which goes to zero

Answer 4

$$ \underbrace{\vphantom{\frac11}+1}_{\text{length }1} \underbrace{-\frac12-\frac13}_{\text{length }2} \underbrace{+\frac14+\frac15+\frac16}_{\text{length }3} \underbrace{-\frac17-\frac18-\frac19-\frac1{10}}_{\text{length }4} \underbrace{+\frac1{11}+\frac1{12}+\frac1{13}+\frac1{14}+\frac1{15}}_{\text{length }5}-\ldots $$ The absolute values of the terms of the same-sign block of length $n$ are from $$ \dfrac1{n(n-1)/2+1}\quad\text{to}\quad\dfrac1{n(n+1)/2} $$ and the sum of the block must satisfy $$ \frac2{n+1}=\frac{n}{n(n+1)/2}\le(-1)^{n-1}\text{sum}\le\frac{n}{n(n-1)/2+1}\lt\frac2{n-1} $$ which tends to $0$.

The absolute value of the sum of the same-sign block of length $n$ and the same-sign block of length $n+1$ is at most $$ \frac2{n-1}-\frac2{n+2}=\frac6{(n+2)(n-1)} $$ Thus, the sum of pairs of blocks converge absolutely and the blocks converge to $0$. Thus, the full series converges.

Answer 5

To elaborate on Ross Millikan's answer and André Nicolas's comment, denote the $n$-th block sum (ignoring the sign) by $B_n$. Then \begin{align} B_n &= \sum_{i=1}^n \frac1{k_n+n},\\ k_n &= \frac{n(n-1)}{2},\\ k_{n+1} &= k_n+n = \frac{n(n+1)}{2} \le n^2.\tag{1} \end{align} Therefore \begin{align} B_n - B_{n+1} &= \sum_{i=1}^n \frac1{k_n+i} - \sum_{i=1}^{n+1} \frac1{k_{n+1}+i}\\ &= \sum_{i=1}^n \frac1{k_n+i} - \sum_{i=1}^n \frac1{k_{n+1}+i} - \frac1{k_n+2n+1}\\ &= \sum_{i=1}^n \frac{n}{(k_n+i)(k_{n+1}+i)} - \frac1{k_n+2n+1}\ \text{ by } (1)\\ &\ge \sum_{i=1}^n \frac{n}{(k_n+n)(k_n+2n)} - \frac1{k_n+2n+1}\\ &= \frac{n^2}{(k_n+n)(k_n+2n)} - \frac1{k_n+2n+1}\\ &\ge \frac{1}{k_n+2n} - \frac1{k_n+2n+1}\ \text{ by } (1)\\ &> 0. \end{align} Hence the block sum is indeed monotonic decreasing and the alternating series test applies.

Answer 6

I figured I'd take a different approach to it, by providing upper and lower bounds.

For an upper bound, we'll assume that all values in the positive sequences are the largest of them, and all values in the negative sequences are the smallest. That is, for instance...

$\frac{1}{4} + \frac{1}{5} + \frac{1}{6} < \frac{3}{4}$

And

$\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}>\frac{4}{10} = \frac{2}{5}$

For the positive terms, we have that the sums are bounded above by $\frac{2n+1}{n(2n+1)+1} < \frac{1}{n}$, while the negative terms' sums are bounded below by $\frac{2}{2n+1} > \frac{1}{n+1}$. As such, we can say easily that the series cannot be larger than

$$ 1+\sum_{n=1}^\infty \left(\frac{1}{n} - \frac{1}{n+1}\right) = 2 $$ Similarly, the positive sums are bounded below by $\frac{1}{n+1}$ and the negative sums are bounded above by $\frac{2n}{n(2n-1)+1} < \frac{1}{n-1}$. And so, the series cannot be smaller than

$$ \frac{1}{2}+\sum_{n=2}^\infty \left(\frac{1}{n+1} - \frac{1}{n-1}\right) = -1 $$

As such, we know that the series must be less than 2 and greater than -1, so we know that the series does not diverge, and thus we only need to show that it does not fail to converge due to oscillation. As both the positive and negative sum terms behave as $1/n$ in the limit as $n \to \infty$ (thereby preventing cyclic behaviour), we know that the series must converge.